《Interest Rate Risk Modeling》第四章：M-absolute 和 M-square 风险度量

M-absolute 和 M-square 风险度量

思维导图

从第四章开始比较难了

$M^A$ 和 $M^2$ 控制了组合预期变化的下限

两个重要不等式的推导

首先有

\[V_0 = \sum_{t=t_1}^{t_n} CF_t e^{-\int_0^t f(s)ds}\]

令

\[\begin{aligned} V_H &= V_0 e^{\int_0^H f(s)ds}\\ &= \sum_{t=t_1}^{t_n} CF_t e^{-\int_0^t f(s)ds} e^{\int_0^H f(s)ds}\\ &= \sum_{t=t_1}^{t_n} CF_t e^{\int_t^H f(s)ds} \end{aligned}\]

以及

\[V_H^{\prime} = \sum_{t=t_1}^{t_n} CF_t e^{\int_t^H f^{\prime}(s)ds}\]

那么

\[\begin{aligned} \frac{V_H^{\prime} - V_H}{V_H} &= \frac{1}{V_0 e^{\int_0^H f(s)ds}} \sum_{t=t_1}^{t_n} CF_t (e^{\int_t^H f^{\prime}(s)ds} - e^{\int_t^H f(s)ds})\\ &=\frac{1}{V_0}\sum_{t=t_1}^{t_n} CF_t[e^{\int_t^H f(s)ds}(e^{\int_t^H \Delta f(s)ds}-1)]e^{-\int_0^H f(s)ds}\\ &=\frac{1}{V_0}\sum_{t=t_1}^{t_n} CF_te^{-\int_0^t f(s)ds}(e^{\int_t^H \Delta f(s)ds}-1) \end{aligned}\]

记

\[h(t) = \int_t^H \Delta f(s)ds\]

关于 $M^A$ 的不等式

$\Delta f(t)$ 的边界分别是 $K_1$ 和 $K_2$，即 $K_1 \le \Delta f(t) \le K_2$

若 $H>t$ 时

\[h(t) \ge K_1(H-t) = K_1 |t-H|\]

若 $H \le t$ 时

\[h(t) \ge -K_2(t-H) = -K_2|t-H|\]

于是

\[h(t) \ge \min(K_1, -K_2)|t-H|\]

而

\[\begin{aligned} \min (K_1, -K_2) &= -\max(-K_1, K_2)\\ &\ge -\max(|K_1|, |K_2|)\\ &=-K_3 \end{aligned}\]

则

\[h(t) \ge -K_3|t-H|\]

已知

\[e^x - 1 \ge x\]

那么

\[\begin{aligned} \frac{V_H^{\prime} - V_H}{V_H} & =\frac{1}{V_0}\sum_{t=t_1}^{t_n} CF_te^{-\int_0^t f(s)ds}(e^{\int_t^H \Delta f(s)ds}-1)\\ & \ge \frac{1}{V_0}\sum_{t=t_1}^{t_n} CF_te^{-\int_0^t f(s)ds}h(t)\\ & \ge \frac{1}{V_0}\sum_{t=t_1}^{t_n} CF_te^{-\int_0^t f(s)ds} (-K_3|t-H|)\\ & = -K_3M^A \end{aligned}\]

关于 $M^2$ 的不等式

记

\[\frac{d\Delta f(t)}{dt} = g(t) \le K_4\]

那么

\[\begin{aligned} h(t) &= \int_t^H \Delta f(s)ds\\ & = t\Delta f(t)|_{t}^{H} - \int_t^H s g(s)ds\\ & = H\Delta f(H) - t\Delta f(t) - \int_t^H s g(s)ds\\ & = (H-t)\Delta f(H) + t\Delta f(H) - t\Delta f(t) - \int_t^H s g(s)ds\\ & = (H-t)\Delta f(H) + t\int_t^H g(s)ds - \int_t^H s g(s)ds\\ & = (H-t)\Delta f(H) + \int_t^H(t-s) g(s)ds\\ \end{aligned}\]

若 $H>t$ 时

\[\begin{aligned} \int_t^H (t-s) g(s)ds & \ge \int_t^H (t-s) K_4ds\\ &=-K_4(t-H)^2/2 \end{aligned}\]

若 $H \le t$ 时

\[\begin{aligned} \int_t^H (t-s) g(s)ds & = -\int_H^t (t-s) g(s)ds\\ & \ge -\int_H^t (t-s) K_4ds\\ & = -K_4(t-H)^2/2 \end{aligned}\]

那么

\[h(t) \ge (H-t)\Delta f(H) -K_4(t-H)^2/2\]

已知

\[e^x - 1 \ge x\]

那么

\[\begin{aligned} \frac{V_H^{\prime} - V_H}{V_H} & =\frac{1}{V_0}\sum_{t=t_1}^{t_n} CF_te^{-\int_0^t f(s)ds}(e^{\int_t^H \Delta f(s)ds}-1)\\ & \ge \frac{1}{V_0}\sum_{t=t_1}^{t_n} CF_te^{-\int_0^t f(s)ds}h(t)\\ & \ge \frac{1}{V_0}\sum_{t=t_1}^{t_n} CF_te^{-\int_0^t f(s)ds} \left((H-t)\Delta f(H) -K_4(t-H)^2/2\right)\\ & = (H-D)\Delta f(H) -K_4M^2/2 \end{aligned}\]

凸性效应（CE）和风险效应（RE）的推导

\[\begin{aligned} R(H) &= \frac{V_H^{\prime} - V_0}{V_0}\\ &=\frac{\sum_{t=t_1}^{t_n} CF_t e^{\int_t^H f^{\prime}(s)ds}}{\sum_{t=t_1}^{t_n} CF_t e^{-\int_0^t f(s)ds}} - 1\\ &=\frac{\sum_{t=t_1}^{t_n} CF_t e^{\int_0^H f^{\prime}(s)ds - \int_0^t f^{\prime}(s)ds}}{\sum_{t=t_1}^{t_n} CF_t e^{-\int_0^t f(s)ds}} - 1\\ &=\frac{e^{\int_0^H f^{\prime}(s)ds}\sum_{t=t_1}^{t_n} CF_t e^{-\int_0^t f(s)ds}e^{-\int_0^t \Delta f(s)ds}}{\sum_{t=t_1}^{t_n} CF_t e^{-\int_0^t f(s)ds}} - 1\\ &=\frac{e^{\int_0^H f(s) + \Delta f(s)ds}\sum_{t=t_1}^{t_n} CF_t e^{-\int_0^t f(s)ds}e^{-\int_0^t \Delta f(s)ds}}{\sum_{t=t_1}^{t_n} CF_t e^{-\int_0^t f(s)ds}} - 1\\ &=\frac{e^{\int_0^H f(s)}\sum_{t=t_1}^{t_n} CF_t e^{-\int_0^t f(s)ds}e^{\int_t^H \Delta f(s)ds}}{\sum_{t=t_1}^{t_n} CF_t e^{-\int_0^t f(s)ds}} - 1\\ \end{aligned}\]

令

\[R_F(H) = e^{\int_0^H f(s)ds} - 1\]

记

\[k(t) = e^{\int_t^H \Delta f(s)ds}\]

对 $k(t)$ 在 $H$ 做 Taylor 展开

\[\begin{aligned} k(t) &= e^{\int_t^H \Delta f(s)ds}\\ &= e^{-\int_H^t \Delta f(s)ds}\\ &= k(H) + (t-H)k'(H) + \frac{1}{2}(t-H)^2k''(H) + \varepsilon\\ &= 1 + (t-H)(-\Delta f(H)) + \frac{1}{2}(t-H)^2(\Delta f(H)^2 - \frac{d(\Delta f(t))}{dt}|_{t=H}) + \varepsilon\\ &= 1 + (t-H)(-\Delta f(H)) + \frac{1}{2}(t-H)^2(\Delta f(H)^2 - g(H)) + \varepsilon\\ \end{aligned}\]

代入得到

\[\begin{aligned} R(H) &= R_F(H) + \gamma_1 (D-H) + \gamma_2 M^2 + \varepsilon\\ \gamma_1 &= -\Delta f(H)(1+R_F(H))\\ \gamma_2 &= \frac{1}{2}(1+R_F(H))(\Delta f(H)^2 - g(H))\\ \gamma_2 &= CE - RE\\ CE &= \frac{1}{2}(1+R_F(H))\Delta f(H)^2\\ RE &= \frac{1}{2}(1+R_F(H))g(H) \end{aligned}\]